##### Question 10.6.2

A regular pentagon $ABCDE$ has diagonals $AC$ and $BE$ intersecting at $F$, and $AC$ and $BD$ intersecting at $G$. Show that $\frac{AC}{AG} = \frac{AG}{AF} = \phi$ where $\phi$ is the ‘golden ratio’, $(1+\sqrt5)/2$. You may assume that $AB = AG$, as shown in Question 10.4.2. Show hint